126k views
0 votes
A ball is thrown upwards into the air with an initial velocity of

17.5 m/s. What is the maximum height that the ball traveled
during that time (assuming there is no air resistance). Round your
answer to two decimals places.

User Deltaluca
by
5.7k points

1 Answer

6 votes

Answer:

y = 15.61 [m]

Step-by-step explanation:

To solve this problem we must use the following kinematics equation. We must bear in mind that the acceleration of gravity is negative since the direction of gravity is opposite to the direction of motion.


v_(f) ^(2) = v_(i) ^(2) -(2*g*y)

where:

Vf = final velocity = 0

Vi = initial velocity = 17.5 [m/s]

g = gravity acceleration = 9.81 [m/s^2]

y = height [m]

The final speed is zero, since the ball reaches the maximum height its speed will be zero since it cannot be lifted up more.

0 = (17.5)^2 - (2*9.81*y)

306.25 = 2*9.81*y

y = 15.61 [m]

User Sanjay Rabari
by
7.1k points